Let $T$ be a compact operator on $l^2$. Let $T_n$ be finite rank operators and $T_n \to T$ in the operator norm. Is it true that the eigenvalues and eigenvectors of $T_n$ converge to eigenvalues and eigenvectors of $T$?
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1$\begingroup$ Which eigenvectors of $T_n$ are supposed to converge to eigenvectors of $T$? All of them? Some of them? $\endgroup$– Yemon ChoiCommented Mar 24, 2011 at 10:15
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1$\begingroup$ Presumably you are motivated by the self-adjoint case. Which other cases have you tried, or heard of? $\endgroup$– Yemon ChoiCommented Mar 24, 2011 at 10:18
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$\begingroup$ This looks like a homework problem that was slightly open ended. I vote to close. $\endgroup$– Bill JohnsonCommented Mar 24, 2011 at 11:26
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1$\begingroup$ I am not so sure this is homework; but I think a more precise question would be better received. Perhaps the original question is motivated by particular examples that have extra structure not present for general compact operators on Hilbert space? (E.g. integral kernel, Toeplitz or band structure.) $\endgroup$– Yemon ChoiCommented Mar 24, 2011 at 19:20
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1$\begingroup$ This is not a homework. This comes from expansion of a solution to PDE with Robin boundary condition in some basis. I've obtained an infinite system of equations Av=v, where A is a band matrix, is square summable (entries in k-th row are of the order 1/k) so A is compact operator on $l^2$. In general I want to show that the eigenvector of truncated matrix A is good approximation of the solution to original problem if the truncation rank is large enough. $\endgroup$– SzopaCommented Mar 25, 2011 at 8:25
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2 Answers
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For any compact set K of complex numbers disjoint from the spectrum of T, there is $\epsilon > 0$ such that for every operator S with $\|S-T\| < \epsilon$, K is disjoint from the spectrum of $S$. Namely, you can take $\epsilon = \inf_{\lambda \in K} \|(T-\lambda)^{-1}\|^{-1}$. So the eigenvalues of $T_n$ do converge in that sense to the spectrum of $T$ (not necessarily eigenvalues, because $T$ may not have any).
answered Mar 24, 2011 at 16:10
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1$\begingroup$ If T has nonzero eigenvalues, then convergence of the resolvent and Dunford calculus can be used to show that the spectral projections converge. $\endgroup$ Commented Mar 24, 2011 at 18:19
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For eigenvectors there is no chance. One may approximate the identity map $T$ on $\Bbb R^2$ with a symmetric matrix $T_n$ whose eigenvalues are $1$ and $1-1/n$. The eigenvectors are perpendicular to each other, but otherwise their direction is entirely optional. So by choosing directions erratically one can avoid convergence. Of course some subsequence will converge.
answered Mar 25, 2011 at 10:58